Question

In a Gallup poll, 26.6% of 5000 people reported a Body Mass Index (BMI) greater than 30, which is classified as obese. The 5000 people polled are a random sample of adults, aged 18 and older, from the United States population. What is the 99% confidence interval for the proportion of the population who are obese

Answer 1

Answer:

The 99% confidence interval for the proportion of the population who are obese is (0.25, 0.282)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 5000, \pi = 0.266$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.266 - 2.575\sqrt{\frac{0.266*0.734}{5000}} = 0.25$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.266 + 2.575\sqrt{\frac{0.266*0.734}{5000}} = 0.282$

The 99% confidence interval for the proportion of the population who are obese is (0.25, 0.282)